Ans: C
$\begin{array}{rcl}
(x-8)(x+ \alpha) – 6 & \equiv & (x-9)^2 + \beta \\
x^2 + (\alpha – 8)x + ( -8\alpha – 6) & \equiv & x^2 – 18x + 81 + \beta
\end{array}$
$\begin{array}{rcl}
(x-8)(x+ \alpha) – 6 & \equiv & (x-9)^2 + \beta \\
x^2 + (\alpha – 8)x + ( -8\alpha – 6) & \equiv & x^2 – 18x + 81 + \beta
\end{array}$
By comparing the coefficients of both sides, we have
$\left\{ \begin{array}{ll}
\alpha – 18 = -18 & \ldots \unicode{x2460} \\
-8\alpha – 6 = 81 + \beta & \ldots \unicode{x2461}
\end{array} \right.$
From $\unicode{x2460}$, we have
$\begin{array}{rcl}
\alpha – 8 & = & -18 \\
\alpha & = & -10
\end{array}$
Sub. $\alpha = -10$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
-8(-10) – 6 & = & 81 + \beta \\
\beta & = & -7
\end{array}$
